用lambda连接一系列紧凑、快速且类型安全的数组

@SafeVarargs
public static <T> T[] concat( T[]... arrays ) {
  return( Stream.of( arrays ).reduce( ( arr1, arr2 ) -> {
      T[] rslt = Arrays.copyOf( arr1, arr1.length + arr2.length );
      System.arraycopy( arr2, 0, rslt, arr1.length, arr2.length );
      return( rslt );
    } ).orElse( null ) );
};

在没有参数的情况下调用时返回null

例如，具有3个阵列：

String[] a = new String[] { "a", "b", "c", "d" };
String[] b = new String[] { "e", "f", "g", "h" };
String[] c = new String[] { "i", "j", "k", "l" };

concat( a, b, c );  // [a, b, c, d, e, f, g, h, i, j, k, l]

“……可能是唯一通用和类型安全的方法”–适用于：

Number[] array1 = { 1, 2, 3 };
Number[] array2 = { 4.0, 5.0, 6.0 };
Number[] array = concat( array1, array2 );  // [1, 2, 3, 4.0, 5.0, 6.0]

这里是silvertab解决方案的一个修改，对泛型进行了改进：

static <T> T[] concat(T[] a, T[] b) {
    final int alen = a.length;
    final int blen = b.length;
    final T[] result = (T[]) java.lang.reflect.Array.
            newInstance(a.getClass().getComponentType(), alen + blen);
    System.arraycopy(a, 0, result, 0, alen);
    System.arraycopy(b, 0, result, alen, blen);
    return result;
}

注意：请参阅Joachim的Java 6解决方案答案。它不仅消除了警告；它也更短，更高效，更容易阅读！

使用Java API：

String[] f(String[] first, String[] second) {
    List<String> both = new ArrayList<String>(first.length + second.length);
    Collections.addAll(both, first);
    Collections.addAll(both, second);
    return both.toArray(new String[both.size()]);
}

你可以试试这个

 public static Object[] addTwoArray(Object[] objArr1, Object[] objArr2){
    int arr1Length = objArr1!=null && objArr1.length>0?objArr1.length:0;
    int arr2Length = objArr2!=null && objArr2.length>0?objArr2.length:0;
    Object[] resutlentArray = new Object[arr1Length+arr2Length]; 
    for(int i=0,j=0;i<resutlentArray.length;i++){
        if(i+1<=arr1Length){
            resutlentArray[i]=objArr1[i];
        }else{
            resutlentArray[i]=objArr2[j];
            j++;
        }
    }

    return resutlentArray;
}

你可以键入你的数组！！！

public int[] mergeArrays(int [] a, int [] b) {
    int [] merged = new int[a.length + b.length];
    int i = 0, k = 0, l = a.length;
    int j = a.length > b.length ? a.length : b.length;
    while(i < j) {
        if(k < a.length) {
            merged[k] = a[k];
            k++;
        }
        if((l - a.length) < b.length) {
            merged[l] = b[l - a.length];
            l++;
        }
        i++;
    }
    return merged;
}

我认为泛型的最佳解决方案是：

/* This for non primitive types */
public static <T> T[] concatenate (T[]... elements) {

    T[] C = null;
    for (T[] element: elements) {
        if (element==null) continue;
        if (C==null) C = (T[]) Array.newInstance(element.getClass().getComponentType(), element.length);
        else C = resizeArray(C, C.length+element.length);

        System.arraycopy(element, 0, C, C.length-element.length, element.length);
    }

    return C;
}

/**
 * as far as i know, primitive types do not accept generics 
 * http://stackoverflow.com/questions/2721546/why-dont-java-generics-support-primitive-types
 * for primitive types we could do something like this:
 * */
public static int[] concatenate (int[]... elements){
    int[] C = null;
    for (int[] element: elements) {
        if (element==null) continue;
        if (C==null) C = new int[element.length];
        else C = resizeArray(C, C.length+element.length);

        System.arraycopy(element, 0, C, C.length-element.length, element.length);
    }
    return C;
}

private static <T> T resizeArray (T array, int newSize) {
    int oldSize =
            java.lang.reflect.Array.getLength(array);
    Class elementType =
            array.getClass().getComponentType();
    Object newArray =
            java.lang.reflect.Array.newInstance(
                    elementType, newSize);
    int preserveLength = Math.min(oldSize, newSize);
    if (preserveLength > 0)
        System.arraycopy(array, 0,
                newArray, 0, preserveLength);
    return (T) newArray;
}